SOLUTION: Please lend me a helping hand. I would really appreciate your help and may God bless you for your continued services.
1)What are the factors of x3 + 6x2 - x - 30?
a. (x + 1)(x
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Question 48455: Please lend me a helping hand. I would really appreciate your help and may God bless you for your continued services.
1)What are the factors of x3 + 6x2 - x - 30?
a. (x + 1)(x - 2)(x + 15)
b. (x + 1)(x + 3)(x - 10)
c. (x - 2)(x + 3)(x + 5)
d. (x - 1)(x + 5)(x + 6)
2)What are all the possible rational zeros for f(x) = x3 + 8x + 6?
a. +1, +6
b. +1, +2, +6
c. +1, +2, +3, +6
d. +1, +2, +3, +6, +12
3)What are all the possible rational zeros for f(x) = 6x3 + 6x2 - 15x - 2?
a. +1, +2, +1/2, +1/3, +2/3, +1/6
b. +1, +2, +1/3, +2/3
c. +1, +2, +3,
d. +1, +2, +3, +6,
4)Identify all rational zeros for f(x) = x3 + 4x2 - 3x - 18.
a. 1, -3, -6
b. -1, 2, -9
c. 2, -3, -3
d. -1, -3, 6
5)Identify all the rational zeros for f(x) = x3 - x2 - 8x + 12.
a. 2, 2, -3
b. 2, -2, 3
c. -1, 3, 4
d. 1, -3, 4
6)Identify all the rational zeros for f(x) = x3 + 4x2 - 2x + 15.
a. 1
b. 3
c. -5
d. 1, 3, -5
7)
How many positive real zeros does the funtion f(x) = x4 + x3 - 7x - 1 have?
a. 3 or 1
b. 1
c. 2 or 0
d. 0
8)How many negative real zeros does the function f(x) = x5 + x4 - x2 - x + 1 have?
a. 4, 2, or 0
b. 3
c. 2 or 0
d. 3 or 1
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1)What are the factors of x^3 + 6x^2 - x - 30?
c. (x - 2)(x + 3)(x + 5)
2)What are all the possible rational zeros for f(x) = x^3 + 8x + 6?
c. +1, +2, +3, +6
3)What are all the possible rational zeros for f(x) = 6x^3 + 6x^2 - 15x - 2?
a. +1, +2, +1/2, +1/3, +2/3, +1/6
4)Identify all rational zeros for f(x) = x^3 + 4x^2 - 3x - 18.
c. 2, -3, -3
5)Identify all the rational zeros for f(x) = x^3 - x^2 - 8x + 12.
a. 2, 2, -3
6)Identify all the rational zeros for f(x) = x^3 + 4x^2 - 2x + 15.
c. -5
7)How many positive real zeros does the funtion f(x) = x^4 + x^3 - 7x - 1 have?
b. 1
8)How many negative real zeros does the function f(x) = x^5 + x^4 - x^2 - x + 1 have?
d. 3 or 1
Cheers,
Stan H.
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